To1in 的BUU刷题记录——4
9.强网杯2019 Copperstudy
nc连上去第0层是爆sha256，我自己写的比较丑，就不贴了。
第1层
[+]Generating challenge 1
[+]n=13112061820685643239663831166928327119579425830632458568801544406506769461279590962772340249183569437559394200635526183698604582385769381159563710823689417274479549627596095398621182995891454516953722025068926293512505383125227579169778946631369961753587856344582257683672313230378603324005337788913902434023431887061454368566100747618582590270385918204656156089053519709536001906964008635708510672550219546894006091483520355436091053866312718431318498783637712773878423777467316605865516248176248780637132615807886272029843770186833425792049108187487338237850806203728217374848799250419859646871057096297020670904211
[+]e=3
[+]m=random.getrandbits(512)
[+]c=pow(m,e,n)=15987554724003100295326076036413163634398600947695096857803937998969441763014731720375196104010794555868069024393647966040593258267888463732184495020709457560043050577198988363754703741636088089472488971050324654162166657678376557110492703712286306868843728466224887550827162442026262163340935333721705267432790268517
[+]((m>>72)<<72)=2519188594271759205757864486097605540135407501571078627238849443561219057751843170540261842677239681908736
[-]long_to_bytes(m).encode('hex')=
看到e=3，n又这么大，虽然不知道给的m的信息有什么用，但是可以直接gmpy2开三次根号，RSA的低指数加密
import gmpy2
from Crypto.Util.number import *
c=15987554724003100295326076036413163634398600947695096857803937998969441763014731720375196104010794555868069024393647966040593258267888463732184495020709457560043050577198988363754703741636088089472488971050324654162166657678376557110492703712286306868843728466224887550827162442026262163340935333721705267432790268517
m = gmpy2.iroot(c,3)[0]
print(long_to_bytes(m).hex())
第2层
[+]Generating challenge 2
[+]n=12784625729032789592766625203074018101354917751492952685083808825504221816847310910447532133616954262271205877651255598995305639194329607493047941212754523879402744065076183778452640602625242851184095546100200565113016690161053808950384458996881574266573992526357954507491397978278604102524731393059303476350167738237822647246425836482533150025923051544431330502522043833872580483142594571802189321599016725741260254170793393777293145010525686561904427613648184843619301241414264343057368192416551134404100386155751297424616254697041043851852081071306219462991969849123668248321130382231769250865190227630009181759219
[+]e=65537
[+]m=random.getrandbits(512)
[+]c=pow(m,e,n)=627824086157119245056478875800598959553774250161670787506083253960788230737588761787385686125828765665617567887904228030839535317987589608761534500003128247164233774794784231518212804270056404565710426613938264302998015421153393879729263551292024543756422702956470022959537221269172084619081368498693930550456153543628170306324206266216348386707008661128717431426237486511309767286175518238620230507201952867261283880986868752676549613958785288914989429224582849218395471672295410036858881836363364885164276983237312235831591858044908369376855484127614933545955544787160352042318378588039587911741028067576722790778
[+]((p>>128)<<128)=97522826022187678545924975588711975512906538181361325096919121233043973599759518562689050415761485716705615149641768982838255403594331293651224395590747133152128042950062103156564440155088882592644046069208405360324372057140890317518802130081198060093576841538008960560391380395697098964411821716664506908672
[-]long_to_bytes(m).encode('hex')=
可以看到给了p的相关信息，给出了p的高位，n大约是2048位，那么p估计是1024位，所以应该是已知p高位的攻击。先贴代码，原理贴链接。
n = 12784625729032789592766625203074018101354917751492952685083808825504221816847310910447532133616954262271205877651255598995305639194329607493047941212754523879402744065076183778452640602625242851184095546100200565113016690161053808950384458996881574266573992526357954507491397978278604102524731393059303476350167738237822647246425836482533150025923051544431330502522043833872580483142594571802189321599016725741260254170793393777293145010525686561904427613648184843619301241414264343057368192416551134404100386155751297424616254697041043851852081071306219462991969849123668248321130382231769250865190227630009181759219
p_fake = 97522826022187678545924975588711975512906538181361325096919121233043973599759518562689050415761485716705615149641768982838255403594331293651224395590747133152128042950062103156564440155088882592644046069208405360324372057140890317518802130081198060093576841538008960560391380395697098964411821716664506908672
pbits = 1024
kbits = 128
pbar = p_fake & (2^pbits-2^kbits)
PR.<x> = PolynomialRing(Zmod(n))
f = x + pbar
x0 = f.small_roots(X=2^kbits, beta=0.4)[0] # find root < 2^kbits with factor >= n^0.3
print (x0 + pbar)
p=x0 + pbar
q=n//p
然后有了p、q就可以RSA解密了。
p=97522826022187678545924975588711975512906538181361325096919121233043973599759518562689050415761485716705615149641768982838255403594331293651224395590747133152128042950062103156564440155088882592644046069208405360324372057140890317518802130081198060093576841538008960560661715295741651653499691458486798196487
n=12784625729032789592766625203074018101354917751492952685083808825504221816847310910447532133616954262271205877651255598995305639194329607493047941212754523879402744065076183778452640602625242851184095546100200565113016690161053808950384458996881574266573992526357954507491397978278604102524731393059303476350167738237822647246425836482533150025923051544431330502522043833872580483142594571802189321599016725741260254170793393777293145010525686561904427613648184843619301241414264343057368192416551134404100386155751297424616254697041043851852081071306219462991969849123668248321130382231769250865190227630009181759219
q=n//p
c=627824086157119245056478875800598959553774250161670787506083253960788230737588761787385686125828765665617567887904228030839535317987589608761534500003128247164233774794784231518212804270056404565710426613938264302998015421153393879729263551292024543756422702956470022959537221269172084619081368498693930550456153543628170306324206266216348386707008661128717431426237486511309767286175518238620230507201952867261283880986868752676549613958785288914989429224582849218395471672295410036858881836363364885164276983237312235831591858044908369376855484127614933545955544787160352042318378588039587911741028067576722790778
e=65537
d=inverse(e,(p-1)*(q-1))
print(long_to_bytes(pow(c,d,n)).hex())
第3层
[+]Generating challenge 3
[+]n=92896523979616431783569762645945918751162321185159790302085768095763248357146198882641160678623069857011832929179987623492267852304178894461486295864091871341339490870689110279720283415976342208476126414933914026436666789270209690168581379143120688241413470569887426810705898518783625903350928784794371176183
[+]e=3
[+]m=random.getrandbits(512)
[+]c=pow(m,e,n)=56164378185049402404287763972280630295410174183649054805947329504892979921131852321281317326306506444145699012788547718091371389698969718830761120076359634262880912417797038049510647237337251037070369278596191506725812511682495575589039521646062521091457438869068866365907962691742604895495670783101319608530
[+]d&((1<<512)-1)=787673996295376297668171075170955852109814939442242049800811601753001897317556022653997651874897208487913321031340711138331360350633965420642045383644955
[-]long_to_bytes(m).encode('hex')=
可以看到给出了d的相关信息，应该是 Partial Key Exposure Attack(部分私钥暴露攻击)
def partial_p(p0, kbits, n):
PR.<x> = PolynomialRing(Zmod(n))
nbits = n.nbits()
f = 2^kbits*x + p0
f = f.monic()
roots = f.small_roots(X=2^(nbits//2-kbits), beta=0.3) # find root < 2^(nbits//2-kbits) with factor >= n^0.3
if roots:
x0 = roots[0]
p = gcd(2^kbits*x0 + p0, n)
return ZZ(p)
def find_p(d0, kbits, e, n):
X = var('X')
for k in range(1, e+1):
results = solve_mod([e*d0*X - k*X*(n-X+1) + k*n == X], 2^kbits)
for x in results:
p0 = ZZ(x[0])
p = partial_p(p0, kbits, n)
if p:
return p
if __name__ == '__main__':
n = 92896523979616431783569762645945918751162321185159790302085768095763248357146198882641160678623069857011832929179987623492267852304178894461486295864091871341339490870689110279720283415976342208476126414933914026436666789270209690168581379143120688241413470569887426810705898518783625903350928784794371176183
e = 3
d = 787673996295376297668171075170955852109814939442242049800811601753001897317556022653997651874897208487913321031340711138331360350633965420642045383644955
nbits = n.nbits()
kbits = d.nbits()
print ("lower %d bits (of %d bits) is given" % (kbits, nbits))
p = find_p(d, kbits, e, n)
q = n//p
print ("d0 = %d" % d)
print ("d = %d" % inverse_mod(e, (p-1)*(q-1)))
d有了就走个流程
d=61931015986410954522379841763963945834108214123439860201390512063842165571430799255094107119082046571341221952786658415661511901536119262974324197242727901361853519060099176095718398341546521709753140715090423775413590463159715914497625346364363050316931779727154988269576808476796380941227956316802411370267
c=56164378185049402404287763972280630295410174183649054805947329504892979921131852321281317326306506444145699012788547718091371389698969718830761120076359634262880912417797038049510647237337251037070369278596191506725812511682495575589039521646062521091457438869068866365907962691742604895495670783101319608530
n=92896523979616431783569762645945918751162321185159790302085768095763248357146198882641160678623069857011832929179987623492267852304178894461486295864091871341339490870689110279720283415976342208476126414933914026436666789270209690168581379143120688241413470569887426810705898518783625903350928784794371176183
print(long_to_bytes(pow(c,d,n)).hex())
https://weichujian.github.io/2020/05/27/rsa%E5%B7%B2%E7%9F%A5%E9%AB%98%E4%BD%8D%E6%94%BB%E5%87%BB1/
第4层
[+]Generating challenge 4
[+]e=3
[+]m=random.getrandbits(512)
[+]n1=78642188663937191491235684351005990853149481644703243255021321296087539054265733392095095639539412823093600710316645130404423641473150336492175402885270861906530337207734106926328737198871118125840680572148601743121884788919989184318198417654263598170932154428514561079675550090698019678767738203477097731989
[+]c1=pow(m,e,n1)=23419685303892339080979695469481275906709035609088426118328601771163101123641599051556995351678670765521269546319724616458499631461037359417701720430452076029312714313804716888119910334476982840024696320503747736428099717113471541651211596481005191146454458591558743268791485623924245960696651150688621664860
[+]n2==98174485544103863705821086588292917749386955237408645745685476234349659452606822650329076955303471252833860010724515777826660887118742978051231030080666542833950748806944312437614585352818344599399156268450521239843157288915059003487783576003027303399985723834248634230998110618288843582573006048070816520647
[+]c2=pow(m,e,n2)=72080679612442543693944655041130370753964497034378634203383617624269927191363529233872659451561571441107920350406295389613006330637565645758727103723546610079332161151567096389071050158035757745766399510575237344950873632114050632573903701015749830874081198250578516967517980592506626547273178363503100507676
[+]n3=91638855323231795590642755267985988356764327384001022396221901964430032527111968159623063760057482761918901490239790230176524505469897183382928646349163030620342744192731246392941227433195249399795012672172947919435254998997253131826888070173526892674308708289629739522194864912899817994807268945141349669311
[+]c3=pow(m,e,n3)=22149989692509889061584875630258740744292355239822482581889060656197919681655781672277545701325284646570773490123892626601106871432216449814891757715588851851459306683123591338089745675044763551335899599807235257516935037356212345033087798267959242561085752109746935300735969972249665700075907145744305255616
[-]long_to_bytes(m).encode('hex')=
e=3，给了三组n、c，低加密指数广播攻击
import random
import gmpy2
from Crypto.Util.number import *
def broadcast(n1, n2 ,n3, c1, c2, c3):
n = [n1, n2, n3]
C = [c1, c2, c3]
N = 1
for i in n:
N *= i
Ni = []
for i in n:
Ni.append(N // i)
T = []
for i in range(3):
T.append(gmpy2.invert(Ni[i], n[i]))
X = 0
for i in range(3):
X += C[i] * Ni[i] * T[i]
m3 = X % N
m = gmpy2.iroot(m3, 3)
return m[0]
def main():
e = 3
n1 = 78642188663937191491235684351005990853149481644703243255021321296087539054265733392095095639539412823093600710316645130404423641473150336492175402885270861906530337207734106926328737198871118125840680572148601743121884788919989184318198417654263598170932154428514561079675550090698019678767738203477097731989
c1 =23419685303892339080979695469481275906709035609088426118328601771163101123641599051556995351678670765521269546319724616458499631461037359417701720430452076029312714313804716888119910334476982840024696320503747736428099717113471541651211596481005191146454458591558743268791485623924245960696651150688621664860
n2 = 98174485544103863705821086588292917749386955237408645745685476234349659452606822650329076955303471252833860010724515777826660887118742978051231030080666542833950748806944312437614585352818344599399156268450521239843157288915059003487783576003027303399985723834248634230998110618288843582573006048070816520647
c2 = 72080679612442543693944655041130370753964497034378634203383617624269927191363529233872659451561571441107920350406295389613006330637565645758727103723546610079332161151567096389071050158035757745766399510575237344950873632114050632573903701015749830874081198250578516967517980592506626547273178363503100507676
n3 = 91638855323231795590642755267985988356764327384001022396221901964430032527111968159623063760057482761918901490239790230176524505469897183382928646349163030620342744192731246392941227433195249399795012672172947919435254998997253131826888070173526892674308708289629739522194864912899817994807268945141349669311
c3= 22149989692509889061584875630258740744292355239822482581889060656197919681655781672277545701325284646570773490123892626601106871432216449814891757715588851851459306683123591338089745675044763551335899599807235257516935037356212345033087798267959242561085752109746935300735969972249665700075907145744305255616
m = broadcast(n1, n2 ,n3, c1, c2, c3)
print (long_to_bytes(m).hex())
if __name__=="__main__":
main()
第5层
[+]Generating challenge 5
[+]n=113604829563460357756722229849309932731534576966155520277171862442445354404910882358287832757024693652075211204635679309777620586814014894544893424988818766425089667672311645586528776360047956843961901352792631908859388801090108188344342619580661377758180391734771694803991493164412644148805229529911069578061
[+]e=7
[+]m=random.getrandbits(512)
[+]c=pow(m,e,n)=112992730284209629010217336632593897028023711212853788739137950706145189880318698604512926758021533447981943498594790549326550460216939216988828130624120379925895123186121819609415184887470233938291227816332249857236198616538782622327476603338806349004620909717360739157545735826670038169284252348037995399308
[+]x=pow(m+1,e,n)=112992730284209629010217336632593897028023711212853788739137950706145189880318698604512926758021552486915464025361447529153776277710423467951041523831865232164370127602772602643378592695459331174613894578701940837730590029577336924367384969935652616989527416027725713616493815764725131271563545176286794438175
[-]long_to_bytes(m).encode('hex')=
给了两组有关明文的数据，高位相同的m，相关明文攻击。这里服务端给的数据有问题，根据网上的wp改了一下。
def attack(c1, c2, b, e, n):
PR.<x>=PolynomialRing(Zmod(n))
g1 = x^e - c1
g2 = (x+b)^e - c2
def gcd(g1, g2):
while g2:
g1, g2 = g2, g1 % g2
return g1.monic()
return -gcd(g1, g2)[0]
n=113604829563460357756722229849309932731534576966155520277171862442445354404910882358287832757024693652075211204635679309777620586814014894544893424988818766425089667672311645586528776360047956843961901352792631908859388801090108188344342619580661377758180391734771694803991493164412644148805229529911069578061#填写n值
c=16404985139084147094704300764850430964980485772400565266054075398380588297033201409914512724255440373095027298869259036450071617770755361938461322132693877590521575670718076480353565935028734363256919872879837455527948173237810119579078252909879868459848240229599708133153841801633280283847680255816123323196#填写明文加密内容
x=92463268823628386526871956385934776043432833035349654252757452728405540022093349560058649691620353528569690982904353035470935543182784600771655097406007508218346417446808306197613168219068573563402315939576563452451487014381380516422829248470476887447827532913133023890886210295009811931573875721299817276803#填写相关明文加密内容
e=7#填写加密指数，不宜太大，一般不超过100
b=1#填写一次函数相差的常数项
m = attack(c,x,b,e,n)
print(hex(m))
#https://huangx607087.online/2021/01/22/RSA-Notes4/
第6层
[+]Generating challenge 6 [+]n=0xbadd260d14ea665b62e7d2e634f20a6382ac369cd44017305b69cf3a2694667ee651acded7085e0757d169b090f29f3f86fec255746674ffa8a6a3e1c9e1861003eb39f82cf74d84cc18e345f60865f998b33fc182a1a4ffa71f5ae48a1b5cb4c5f154b0997dc9b001e441815ce59c6c825f064fdca678858758dc2cebbc4d27L
[+]d=random.getrandbits(1024*0.270)
[+]e=invmod(d,phin)
[+]hex(e)=0x11722b54dd6f3ad9ce81da6f6ecb0acaf2cbc3885841d08b32abc0672d1a7293f9856db8f9407dc05f6f373a2d9246752a7cc7b1b6923f1827adfaeefc811e6e5989cce9f00897cfc1fc57987cce4862b5343bc8e91ddf2bd9e23aea9316a69f28f407cfe324d546a7dde13eb0bd052f694aefe8ec0f5298800277dbab4a33bbL
[+]m=random.getrandbits(512)
[+]c=pow(m,e,n)=0xe3505f41ec936cf6bd8ae344bfec85746dc7d87a5943b3a7136482dd7b980f68f52c887585d1c7ca099310c4da2f70d4d5345d3641428797030177da6cc0d41e7b28d0abce694157c611697df8d0add3d900c00f778ac3428f341f47ecc4d868c6c5de0724b0c3403296d84f26736aa66f7905d498fa1862ca59e97f8f866cL
[-]long_to_bytes(m).encode('hex')=
e很大，d<N的0.292次方，使用Boneh and Durfee attack
import time
debug = True
strict = False
helpful_only = True
dimension_min = 7
def helpful_vectors(BB, modulus):
nothelpful = 0
for ii in range(BB.dimensions()[0]):
if BB[ii,ii] >= modulus:
nothelpful += 1
print( nothelpful, "/", BB.dimensions()[0], " vectors are not helpful")
def matrix_overview(BB, bound):
for ii in range(BB.dimensions()[0]):
a = ('%02d ' % ii)
for jj in range(BB.dimensions()[1]):
a += '0' if BB[ii,jj] == 0 else 'X'
if BB.dimensions()[0] < 60:
a += ' '
if BB[ii, ii] >= bound:
a += '~'
print( a)
def remove_unhelpful(BB, monomials, bound, current):
if current == -1 or BB.dimensions()[0] <= dimension_min:
return BB
for ii in range(current, -1, -1):
# if it is unhelpful:
if BB[ii, ii] >= bound:
affected_vectors = 0
affected_vector_index = 0
for jj in range(ii + 1, BB.dimensions()[0]):
if BB[jj, ii] != 0:
affected_vectors += 1
affected_vector_index = jj
if affected_vectors == 0:
print( "* removing unhelpful vector", ii)
BB = BB.delete_columns([ii])
BB = BB.delete_rows([ii])
monomials.pop(ii)
BB = remove_unhelpful(BB, monomials, bound, ii-1)
return BB
elif affected_vectors == 1:
affected_deeper = True
for kk in range(affected_vector_index + 1, BB.dimensions()[0]):
if BB[kk, affected_vector_index] != 0:
affected_deeper = False
if affected_deeper and abs(bound - BB[affected_vector_index, affected_vector_index]) < abs(bound - BB[ii, ii]):
print( "* removing unhelpful vectors", ii, "and", affected_vector_index)
BB = BB.delete_columns([affected_vector_index, ii])
BB = BB.delete_rows([affected_vector_index, ii])
monomials.pop(affected_vector_index)
monomials.pop(ii)
BB = remove_unhelpful(BB, monomials, bound, ii-1)
return BB
return BB
def boneh_durfee(pol, modulus, mm, tt, XX, YY):
# substitution (Herrman and May)
PR.<u, x, y> = PolynomialRing(ZZ)
Q = PR.quotient(x*y + 1 - u) # u = xy + 1
polZ = Q(pol).lift()
UU = XX*YY + 1
# x-shifts
gg = []
for kk in range(mm + 1):
for ii in range(mm - kk + 1):
xshift = x^ii * modulus^(mm - kk) * polZ(u, x, y)^kk
gg.append(xshift)
gg.sort()
# x-shifts list of monomials
monomials = []
for polynomial in gg:
for monomial in polynomial.monomials():
if monomial not in monomials:
monomials.append(monomial)
monomials.sort()
# y-shifts (selected by Herrman and May)
for jj in range(1, tt + 1):
for kk in range(floor(mm/tt) * jj, mm + 1):
yshift = y^jj * polZ(u, x, y)^kk * modulus^(mm - kk)
yshift = Q(yshift).lift()
gg.append(yshift) # substitution
# y-shifts list of monomials
for jj in range(1, tt + 1):
for kk in range(floor(mm/tt) * jj, mm + 1):
monomials.append(u^kk * y^jj)
# construct lattice B
nn = len(monomials)
BB = Matrix(ZZ, nn)
for ii in range(nn):
BB[ii, 0] = gg[ii](0, 0, 0)
for jj in range(1, ii + 1):
if monomials[jj] in gg[ii].monomials():
BB[ii, jj] = gg[ii].monomial_coefficient(monomials[jj]) * monomials[jj](UU,XX,YY)
if helpful_only:
BB = remove_unhelpful(BB, monomials, modulus^mm, nn-1)
nn = BB.dimensions()[0]
if nn == 0:
print( "failure")
return 0,0
if debug:
helpful_vectors(BB, modulus^mm)
det = BB.det()
bound = modulus^(mm*nn)
if det >= bound:
print( "We do not have det < bound. Solutions might not be found.")
print( "Try with highers m and t.")
if debug:
diff = (log(det) - log(bound)) / log(2)
print( "size det(L) - size e^(m*n) = ", floor(diff))
if strict:
return -1, -1
else:
print( "det(L) < e^(m*n) (good! If a solution exists < N^delta, it will be found)")
if debug:
matrix_overview(BB, modulus^mm)
if debug:
print( "optimizing basis of the lattice via LLL, this can take a long time")
BB = BB.LLL()
if debug:
print( "LLL is done!")
if debug:
print( "looking for independent vectors in the lattice")
found_polynomials = False
for pol1_idx in range(nn - 1):
for pol2_idx in range(pol1_idx + 1, nn):
PR.<w,z> = PolynomialRing(ZZ)
pol1 = pol2 = 0
for jj in range(nn):
pol1 += monomials[jj](w*z+1,w,z) * BB[pol1_idx, jj] / monomials[jj](UU,XX,YY)
pol2 += monomials[jj](w*z+1,w,z) * BB[pol2_idx, jj] / monomials[jj](UU,XX,YY)
PR.<q> = PolynomialRing(ZZ)
rr = pol1.resultant(pol2)
if rr.is_zero() or rr.monomials() == [1]:
continue
else:
print( "found them, using vectors", pol1_idx, "and", pol2_idx)
found_polynomials = True
break
if found_polynomials:
break
if not found_polynomials:
print( "no independant vectors could be found. This should very rarely happen...")
return 0, 0
rr = rr(q, q)
soly = rr.roots()
if len(soly) == 0:
print( "Your prediction (delta) is too small")
return 0, 0
soly = soly[0][0]
ss = pol1(q, soly)
solx = ss.roots()[0][0]
return solx, soly
def example():
N = 0xbadd260d14ea665b62e7d2e634f20a6382ac369cd44017305b69cf3a2694667ee651acded7085e0757d169b090f29f3f86fec255746674ffa8a6a3e1c9e1861003eb39f82cf74d84cc18e345f60865f998b33fc182a1a4ffa71f5ae48a1b5cb4c5f154b0997dc9b001e441815ce59c6c825f064fdca678858758dc2cebbc4d27L
e = 0x11722b54dd6f3ad9ce81da6f6ecb0acaf2cbc3885841d08b32abc0672d1a7293f9856db8f9407dc05f6f373a2d9246752a7cc7b1b6923f1827adfaeefc811e6e5989cce9f00897cfc1fc57987cce4862b5343bc8e91ddf2bd9e23aea9316a69f28f407cfe324d546a7dde13eb0bd052f694aefe8ec0f5298800277dbab4a33bbL
c = 0xe3505f41ec936cf6bd8ae344bfec85746dc7d87a5943b3a7136482dd7b980f68f52c887585d1c7ca099310c4da2f70d4d5345d3641428797030177da6cc0d41e7b28d0abce694157c611697df8d0add3d900c00f778ac3428f341f47ecc4d868c6c5de0724b0c3403296d84f26736aa66f7905d498fa1862ca59e97f8f866cL
delta = .291 # this means that d < N^delta
m = 4 # size of the lattice (bigger the better/slower)
t = int((1-2*delta) * m) # optimization from Herrmann and May
X = 2*floor(N^delta) # this _might_ be too much
Y = floor(N^(1/2)) # correct if p, q are ~ same size
P.<x,y> = PolynomialRing(ZZ)
A = int((N+1)/2)
pol = 1 + x * (A + y)
if debug:
print( "=== checking values ===")
print( "* delta:", delta)
print( "* delta < 0.292", delta < 0.292)
print( "* size of e:", int(log(e)/log(2)))
print( "* size of N:", int(log(N)/log(2)))
print( "* m:", m, ", t:", t)
if debug:
print( "=== running algorithm ===")
start_time = time.time()
solx, soly = boneh_durfee(pol, e, m, t, X, Y)
if solx > 0:
print( "=== solution found ===")
if False:
print( "x:", solx)
print( "y:", soly)
d = int(pol(solx, soly) / e)
m = pow(c,d,N)
print( '[-]d is ' + str(d))
print( '[-]m is: ' + str(m))
print( '[-]hex(m) is: ' + '{:x}'.format(int(m)))
else:
print( "[!]no solution was found!")
print( '[!]All Done!')
if debug:
print(("[!]Timer: %s s" % (time.time() - start_time)))
print( '[!]All Done!')
if __name__ == "__main__":
example()
这道题就先到这里了，不过其中涉及很多RSA的攻击方式还要去深入学习（挖坑，坑++)
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